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3 years ago

When you lift a box, what is the effort force?

Physics

1 answer:

aleksandrvk [35]3 years ago

5 0

Answer:

effort force: The force used to move an object over a distance. resistance force: The force which an effort force must overcome in order to do work on an object via a simple machine. ideal mechanical advantage: The factor by which a mechanism multiplies the force put into it.

Explanation:

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VLD [36.1K]

Answer:O – H

Explanation:

8 0

3 years ago

The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg

kumpel [21]

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

$\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\$

From the graph;

$\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m$

The change in the momentum is;

$\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \ N.s$

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

6 0

2 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

4 0

3 years ago

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles

Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

The right answer is C)

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0

3 years ago

Read 2 more answers

What is the relationship between electrostatic force and electric field?

umka2103 [35]

Electrostatic attraction. The stronger the electric field, the greater the electrostatic force.

7 0

3 years ago